1. Sin cos Formula

how do i get this : [ 2 sin h/2 cos (x + h/2) ] / h ???

2. Originally Posted by silverfox12
how do i get this : [ 2 sin (h/2) cos (x + h/2) ] / h ???
Use:

Sin(A+B) = Sin(A)*Cos(B) + Cos(A)*Sin(B)

Then

Sin(A+B) - Sin(A-B) = ??

replace A = x + h/2 & B = x - h/2

and you are there......

3. Thanks man !

4. I might actually need more help here . As you've said sin(a+b) = sin(a)*cos(b) + cos(a)*sin(b)
Using that : sin(x + h/2) - sin(x - h/2) = [sin(x)*cos(h/2) + cos(x)*sin(h/2)] - [sin(x)*cos(h/2) - cos(x)*sin(h/2)] if got it right . But after some work (namely just adding and subtracting) i get something like this cos(x)*sin(h/2) + cos(x)* sin(h/2)which i think equals 2*sin(h/2)*cos(x) the problem is that according to the textbook it should be like this 2*sin(h/2)* cos(x + h/2) could you please help with this too ? Any idea ?

5. Originally Posted by silverfox12
I might actually need more help here . As you've said sin(a+b) = sin(a)*cos(b) + cos(a)*sin(b)
Using that : sin(x + h/2) - sin(x - h/2) = [sin(x)*cos(h/2) + cos(x)*sin(h/2)] - [sin(x)*cos(h/2) - cos(x)*sin(h/2)] if got it right . But after some work (namely just adding and subtracting) i get something like this cos(x)*sin(h/2) + cos(x)* sin(h/2)which i think equals 2*sin(h/2)*cos(x) the problem is that according to the textbook it should be like this 2*sin(h/2)* cos(x + h/2) could you please help with this too ? Any idea ?
Do the algebra carefully:

sin(x+h) - sin(x) = sin(A+B) - sin(A-B) = 2 * sin(B) * cos(A) = 2 * sin(h/2) * cos(x+h/2)

6. Thanks for the answer and sorry for bothering but i'm really confused. Could you please explain what does A and B represent ? In the last post you wrote A =x + h/2 and B = x - h/2 but then in this one at the last equation you replaced B = h/2 ?

7. Do the algebra carefully (and using pencil + paper instead of staring at the screen) :

sin(x+h) - sin(x)

Substitute A = x + h/2 & B = h/2

= sin(A+B) - sin(A-B)

= 2 * sin(B) * cos(A)

= 2 * sin(h/2) * cos(x+h/2)

First post was an example - not to be used for spoon-feeding.

8. The last question i swear .
I understand this
sin(a+b) - sin(a-b ) = 2cos(a)sin(b)

But how do i determine what should i insert instead of A and B ?

Like in this particular example sin(x + h) - sin( x) . You chose A = x + h/2 and B = h/2 but why ?

sin(x + h/2) - sin(x - h/2)
sin(A + B ) - sin(A - B )
here shouldn't A be equal to x and B equal to h/2 given these formulas ??
I'm getting something wrong just cant find out what exactly . But why do i chose A as the whole(x + h/2) and B only as the h/2 part of (x - h/2); What's the rule of choosing A and B ?

9. Originally Posted by silverfox12
The last question i swear .
I understand this
sin(a+b) - sin(a-b ) = 2cos(a)sin(b)

But how do i determine what should i insert instead of A and B ?

Like in this particular example sin(x + h) - sin( x) . You chose A = x + h/2 and B = h/2 but why ?

sin(x + h/2) - sin(x - h/2)
sin(A + B ) - sin(A - B )
here shouldn't A be equal to x and B equal to h/2 given these formulas ??
I'm getting something wrong just cant find out what exactly . But why do i chose A as the whole(x + h/2) and B only as the h/2 part of (x - h/2); What's the rule of choosing A and B ?
You have the expression you are working with
sin(x+h) - sin(x)
You look to where you want to go
sin(h/2) cos(x+h/2)

The sin(s1) * cos(s2) looks likes a sin(a+b)$\pm$ sin(a-b). [Had it been a cos(s1) cos(s2) or sin(s1) sin(s2), we would use cos(a+b) $\pm$ cos(a-b)]. For convenience choose the minus sign and thus we have
sin(a+b) - sin(a-b) = 2 sin(b) cos(a) = 2 sin(h/2) cos(x+h/2)
So
b = h/2
and
a = x+h/2
Check:
a+b = x + h
a-b = x
which is
sin(x+h) - sin(x)
and what we wanted.

10. Thanks i understand now . (a+b) = x+ h and (a-b) = x
so a = x + b and that we plug into (a+b) = x+h and we get 2b = h - > b = h/2 -> a = x +h/2
Wish you good health for the help .